Concept:
Two planes are parallel if their normal vectors are parallel.
The given plane is
\[
x+3y+5z+1=0
\]
So any plane parallel to it will have the form
\[
x+3y+5z+d=0
\]
where \(d\) is a constant.
Step 1: Use the distance formula.
The distance of the plane
\[
ax+by+cz+d=0
\]
from the origin \((0,0,0)\) is
\[
\frac{|d|}{\sqrt{a^2+b^2+c^2}}
\]
Here,
\[
a=1,\quad b=3,\quad c=5
\]
Therefore,
\[
\sqrt{a^2+b^2+c^2}
=
\sqrt{1^2+3^2+5^2}
\]
\[
=\sqrt{1+9+25}
\]
\[
=\sqrt{35}
\]
Step 2: Given distance is \(5\).
\[
\frac{|d|}{\sqrt{35}}=5
\]
\[
|d|=5\sqrt{35}
\]
Therefore,
\[
d=\pm 5\sqrt{35}
\]
Step 3: Select the answer according to the given options.
The option marked in the paper is
\[
d=5\sqrt{35}
\]
Hence the required plane is
\[
x+3y+5z+5\sqrt{35}=0
\]
Step 4: Final answer.
\[
\boxed{x+3y+5z+5\sqrt{35}=0}
\]